Calculate the pressure that a certain amount of water applies

• MartaBraga
In summary, the conversation is about calculating the pressure exerted by a certain amount of water on the wall of a dam. The equation Pressure = P(atmosphere) + (density(water).height.gravity) is correct for finding the pressure at a certain depth, and it varies linearly. The force on the wall can be calculated by integrating the pressure distribution or by finding the area of the triangle. The atmospheric pressure can be left out as it acts on both sides of the dam and cancels out. The pressure at the bottom is equal to the pressure at the top, and the average pressure can be found by dividing the bottom pressure by 2. The correct formula for the pressure at a certain depth is P = (density

MartaBraga

How do I calculate the pressure that a certain amount of water applies on the wall of a dem(example)?

Is the equation Pressure = P(atmosphere) + (density(water).hight.gravity) correct? Or does that apply only for the pressure aplied on the bottom of the system?

That equation is correct. Realize that "height" is really depth below the water surface.

.You have to integrate the pressure distribution going up the wall...or you could be clever and just find the area of the triangle and do it geometrically.

Thats the correct formula for the pressure at that depth.
(the force sideways is the same as the force downwards because it is a fluid, so pushes in all directions)

At the bottom is the pressure you calculated, at the surface there is obviously no pressure and half way down is half the pressure.

Cyrus said:
.You have to integrate the pressure distribution going up the wall...or you could be clever and just find the area of the triangle and do it geometrically.

How do I do that, I mean, calculate it through the area of the triangle? (I rather do anything than integrate!)

The pressuer at the bottom is: $$P=\rho * g*h$$. The pressure at the top is zero. It varies linearly. I'll leave it to you to show us the rest of the derivation and logic.

Ok, so is it that simple?
I calculated the pressure through that formula and then divided it by 2 and multiplied it by the area... and the result is correct... But I thought I had to use the atmospheric pressure in the formula...

Or is this pure luck?

Can you show your steps. It would be more insightful if you knew why your answer is correct. Right now, you are just blindly plugging away and hoping for the best to come out.

Ok, the problem is to calculate the pressure the water applies on a dem, with 40m of heigth and 150m of lenght.
I calculated the pressure = 1000 (the density) x 40 (the height) x 9,8 (the aceleration of gravity. The result is 392000, which I divided by 2 = 196000. I multiplied this value by the area of the dem in contact with the water, 6000m2. The result is 1,176x10^9.
This was the result I was looking for.

Ok, first of all the water does not exert a 'pressure' on the dam. It has a pressure distribution. This causes a net force acting at the center of pressure. (Think center of gravity, same concept different name).

Yes that exactly right. Do you know what you are doing though?

That's the problem. I just thought, well, if the pressure on the top is 0, and on the bottom is, in this case, 392000, let's find the average pressure, and that's why I divided it by 2. and then, I multiplied it by the area.

(I'm really sorry about the grammar errors, but I'm Portuguese, so there are some words and expressions I don't fully know how to spell or use)

"But I thought I had to use the atmospheric pressure in the formula..."

You can leave out the atmospheric pressure because it is applied to the downstream side of the dam also - hence it cancels out. If there were no water in the lake at all, just the air on both sides, there would be no net force on the dam.

If the problem is to calculate the net force that the water exerts on the dam, then you must include atmospheric pressure. (But as gmax137 states, for some purposes you can forget atmospheric pressure when it acts on both sides of the dam.)

MartaBraga said:
That's the problem. I just thought, well, if the pressure on the top is 0, and on the bottom is, in this case, 392000, let's find the average pressure, and that's why I divided it by 2. and then, I multiplied it by the area.

(I'm really sorry about the grammar errors, but I'm Portuguese, so there are some words and expressions I don't fully know how to spell or use)

Hint: The area of a triangle is 1/2*(base)*(height)!

1. How do you calculate the pressure of water?

To calculate the pressure of water, you will need to know the density of water, which is 1000 kg/m3, and the depth of the water. Use the formula P = ρgh, where P is the pressure, ρ is the density, g is the acceleration due to gravity (9.8 m/s2), and h is the depth of the water in meters.

2. What unit is used to measure water pressure?

The unit commonly used to measure water pressure is Pascal (Pa), which is equivalent to one Newton per square meter (N/m2). Other units that may be used include pounds per square inch (psi) and kilopascal (kPa).

3. How does the amount of water affect the pressure?

The amount of water does not directly affect the pressure. The pressure is determined by the depth of the water and the gravitational force acting on it. However, the volume of water can indirectly affect the pressure if it changes the depth or shape of the water body.

4. Does the temperature of water affect its pressure?

Yes, the temperature of water can affect its pressure. As water warms, it expands and becomes less dense, which can lead to a decrease in pressure. On the other hand, as water cools, it contracts and becomes more dense, which can result in an increase in pressure.

5. Can you calculate the pressure of water in a closed container?

Yes, you can calculate the pressure of water in a closed container using the same formula as before, but with the addition of the atmospheric pressure. The formula will be P = ρgh + Patm, where Patm is the atmospheric pressure. Atmospheric pressure can vary, but it is typically around 1013 hPa or 14.7 psi at sea level.